设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…). 证明:fn(x)=1/(n-1)!∫0xf0(t)(x-t)n-1dt(n=1,2,…);

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问题 设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…).
证明:fn(x)=1/(n-1)!∫0xf0(t)(x-t)n-1dt(n=1,2,…);

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答案n=1时,f1(x)=∫0x0(t)dt,等式成立; 设n=k时,fk(x)=[*]∫0xf0(t)(x-t)k-1dt, 则n=k+1时, fk+1(x)=∫0xfk(t)dt=∫0xdt∫0t[*]f0(u)(t-u)k-1du =[*]∫0xdy∫uxf0(u)(t-u)k-1dt=1/k!∫0xf0(u)*(x-u)kdu 由归纳法得fn(x)=[*]∫0xf0(t)(x-t)n-1dt(7n=1,2,…).

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